Thermodynamics Heat & Work The First Law of Thermodynamics

Transcription

1 Thermodynamics Heat & Work The First Law of Thermodynamics Lana Sheridan De Anza College April 26, 2016

2 Last time applying the ideal gas equation thermal energy heat capacity phase changes latent heat

3 Overview more about phase changes work, heat, and the first law of thermodynamics P-V diagrams applying the first law in various cases heat transfer

4 Evaporation evaporation the process by which a liquid changes to a gas at the liquid surface Since changing from a liquid to a gas requires heat, when a liquid evaporates it takes heat from its surroundings. This is why humans sweat in hot weather, pigs wallow puddles, and dogs pant. All are trying to use evaporation of water to reduce body temperature.

5 Evaporation Ben Franklin noticed that a wet shirt kept him feeling cool on a hot day. He decided to experiment to see if the temperature of objects could be lowered by this process. In 1758 he and John Hadley took a mercury thermometer and repeatedly wet the bulb with ether while using bellows to keep air moving over it. Despite it being a warm day, they recorded temperatures as low as 7 F ( 14 C) at the bulb of the thermometer. This is the basic idea behind refrigeration!

6 Regelation This is a phenomenon seen in water because its density is lower in its solid state. High pressure applied to solid water causes it to melt, even at low temperatures. This makes ice skating work.

7 Phase Change paths

8 Phase Diagrams 1 A typical phase diagram. The dashed green line shows the unusual behavior of water. Diagram by Matthieumarechal, Wikipedia.

9 Liquids below their melting point & above their boiling point Phase changes require the bond structure of the substance to change.

10 Liquids below their melting point & above their boiling point Phase changes require the bond structure of the substance to change. As a liquid is cooled, it will generally start to freeze at one point before another. (Symmetry breaking.) For example, ice cubes in a freezer freeze from the top down. As water freezes solid crystals form and spread throughout. If the water is very pure and cooled without shaking, it can be cooled below 0 C without freezing. This is called supercooling and can happen in some other liquids also under the right conditions.

11 Liquids below their melting point & above their boiling point The same thing can happen when water (or other liquids) are heated just above their boiling points. As the liquid starts to boil, bubbles of vapor form inside it. This happens most easily at defects in the edges of the container. If the container is very smooth and not shaken, the only way the liquid can vaporize is from the middle. Surface tension opposes this. The liquid can be heated a bit above its boiling point. This is called superheating. It is used in bubble chambers to observe charged particles.

12 Liquids below their melting point & above their boiling point Bubble chambers: 1 Left, figure by aarchiba, Wikipedia; right, IOP

13 Heat and Work We now take a closer look at the first law of thermodynamics. To do this, we will take a deeper look at work and heat. We also need to consider our system more carefully.

14 Thermodynamic Equilibrium States We will study thermodynamic systems. These systems are in thermodynamic equilibrium internally. Thermodynamic equilibrium state a state of a system in which every part of the system will be at the same temperature, T, and if the system is a gas, at the same pressure, P. In classical thermodynamics, it is a postulate that any system left isolated will come to a thermal equilibrium state given enough time, and then remain in that state.

15 Thermodynamic Equilibrium States We will study thermodynamic systems. These systems are in thermodynamic equilibrium internally. Thermodynamic equilibrium state a state of a system in which every part of the system will be at the same temperature, T, and if the system is a gas, at the same pressure, P. In classical thermodynamics, it is a postulate that any system left isolated will come to a thermal equilibrium state given enough time, and then remain in that state. In particular, for our present discussion we will be considering an ideal gas. (Can use PV = nrt.)

16 Ideal Gases Ideal gas clarification. We make the following assumptions in the ideal gas model: the volume of the gas particles is negligible compared to the total gas volume molecules are identical hard spheres (will relax this later) collisions between molecules are elastic there are no intermolecular forces (aside from hard-sphere collisions) there are no long-range forces from the environment (can be relaxed) A real gas is behaves as an idea gas when it is at high temperature and low density (far from condensation).

17 Variables The variables we will use can be broken into types: state variables describe system s state / properties T, P, V, and E int. transfer variables describing energy transferred into our out of the system Q, W

18 Variables The variables we will use can be broken into types: state variables describe system s state / properties T, P, V, and E int. transfer variables describing energy transferred into our out of the system Q, W intensive variables variables that don t change value when the system is doubled in size P, T, ρ, c extensive variables variables that double their value when the system is doubled in size V, E int, m, C

19 ergy is entering or leaving ndary Work represents done a change on a gas given state of the system, Imagine compressing a gas in a piston quasi-statically (meaning er variable. slowly In enough this section, so the gas remains in thermal equilibrium). namic systems, work. Work r 7, and here we investigate gas contained in a cylinder e gas occupies a volume V d on the piston. If the pise exerted by the gas on the he force exerted by the pisiston inward and compress e system to remain essenoint of application of the A iston is pushed downward dy d S r 5 dy j^ (Fig. 20.4b), the rk in Chapter 7, 2PA dy is discussion. Because A dy a b work done on the gas as How much work is done Figure on20.4 the gas? Work is done on 1 (20.8) a gas contained in a cylinder at a Figure form Serway & pressure Jewett. P as the piston is P V

20 rvation process ransfer leaving change system, Work done on a gas ection,. Work stigate ylinder lume V the pison the the pismpress essenof the nward b), the P A V dy Definition of work: W = F dr For this system: W = ( F j) dy j = F dy = P A dy = P dv se A dy gas as (20.8) ositive.. If the a Figure 20.4 Work is done on a gas contained in a cylinder at a pressure P as the piston is pushed downward so that the gas is compressed. b Vf W = P dv V i

21 e gas be Work done on a gas fer variservation a process transfer r leaving Here, the volume decreases, so the work done on the gas is positive. a change e system, s section, rk. Work vestigate cylinder volume V f the pisas on the y the piscompress in essenn of the ownward 0.4b), the ause A dy e gas as (20.8) P A a V dy Figure 20.4 Work is done on a gas contained in a cylinder at a pressure P as the piston is b Work done on a gas Vf W = P dv The work done Von a gas i equals the negative of the area under the PV curve. The area The work is negative donehere is the because area under the the P-V curve volume (with is decreasing, the appropriate resulting sign). in positive work. P f P i P V f f V i i V To e ing I dep at e imp of d cur N PV d T i d F par illu

22 The work done on a gas equals the negative of the area under the PV curve. The area P-V diagrams are very is negative useful in here thermodynamics. because volume is decreasing, resulting Example of a P-V diagram: in positive work. Aside: P-V Diagrams P f P i P f depends on the vo at each step of the important graphic of diagram allows curve on a PV diag Notice that the PV diagram. There The work done initial state to a diagram, evalua For the process V particular path tak V f V i illustrate this imp They are graphs offigure pressure 20.5 vs volume A gas is compressed for a fixed quantity (Fig. 20.6). In the (n moles) of an ideal quasi-statically gas. (slowly) from state reduced from V i t i to state f. An outside agent must increases from P i t They contain a lotdo ofpositive information. work on the gas to along this path is 2 1 compress it. Figure form Serway & Jewett, 9th ed, page 602. from P to P at con i

23 Aside: P-V Diagrams Example of a P-V diagram: Pressure i f Volume These diagrams represent the (thermodynamic) equilibrium states of the ideal gas sample, each point is a state, imply the temperature of a known sample, and show the internal energy of the gas. (E int T ) PV = nrt 1 Figure (modified) from Halliday, Resnick, and Walker, page 537.

24 Aside: under P-V the PV curve. Diagrams The area P f P i equals the negative of the area is negative here because the volume is decreasing, resulting Example in positive of work. a P-V diagram: P f i depends on the volume and temper at each step of the process, the state important graphical representation c of diagram allows us to visualize a pr curve on a PV diagram is called the p When Notice these that diagrams the integral include a in path, Equatio they PV diagram. Therefore, we can ident show reversible processes, The eg. compression work done of on gas, a gas in a qua initial show all state theto intermediate a final state is the n diagram, thermal states evaluated passedbetween through, the in show the work done on the gas, For the process of compressing a g particular indicate path the heat taken transferred between to V the in V f V i illustrate the gas. this important point, consi Figure 20.5 A gas is compressed (Fig. 20.6). In the process depicted i quasi-statically (slowly) from state E int = reduced W + Q from V i to V f at constant p i to state f. An outside agent must increases from P i to P f by heating at c do positive work on the gas to along this path is 2P i (V f 2 V i ). In Fig compress it. from P i to P f at constant volume V i an

25 same initial volume, temperature, and pressure, and is assumed to be ideal. In Figure 20.7a, the gas is thermally insulated from its surroundings except at the bottom of Work done depends on the process the gas-filled region, where it is in thermal contact with an energy reservoir. An energy reservoir Different is a source paths of energy or processes that is considered to go from to (V be i, so P great i ) to (V that f, P a finite f ) require transfer of energy to or from the reservoir does not change its temperature. The piston is held different amounts of work. A constant-pressure compression followed by a constant-volume process A constant-volume process followed by a constantpressure compression An arbitrary compression P P P P f f P f f P f f P i V f V i i V P i a b c V f In all of the processes shown above, there are temperature changes during the process. 1 Figure form Serway & Jewett. V i i V P i V f V i i V

26 Heat transfer also depends on the process This process (shown) happens at constant temperature: 20.5 The First Law of The gas is initially at temperature T i. The hand reduces its downward force, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. The g initial tempe T i an conta by a th memb with v above Energy reservoir at T i Energy reservoir at T i a b c Heat Q is transferred to the gas, and the gas does positive work on its surroundings. (Equivalently, negative work is done on the gas.) Figure 20.7 Gas in a cylinder. (a) The gas is in contact with an energy reservoir. The walls of the cylin base in contact with the reservoir is conducting. (b) The gas expands slowly to a larger volume. (c) The half of a volume, with vacuum in the other half. The entire cylinder is perfectly insulating. (d) The gas

27 Heat transfer also depends on the process This process also happens at constant temperature, and has the same start and 20.5 end The First points, Law of (VThermodynamics i, P i ) to f, P f ): 603 The hand reduces its downward force, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. The gas is initially at temperature T i and contained by a thin membrane, with vacuum above. The membrane is broken, and the gas expands freely into the evacuated region. ir at T i No heat is transferred to the gas, and the gas does no work. c (We cannot represent the path for this process on a P-V diagram.) tact with an energy reservoir. The walls of the cylinder are perfectly insulating, but the d

28 First Law of Thermodynamics Reminder: Internal energy, E int or U The energy that a system has as a result of its temperature and all other molecular motions, effects, and configurations, when viewed from a reference frame at rest with respect to the center of mass of the system. 1st Law The change in the internal energy of a system is equal to the sum of the heat added to the system and the work done on the system. E int = W + Q This is just the conservation of energy assuming only the internal energy changes.

29 Applying the 1st Law i W > 0 Volume Some special cases of interest: i Process f Pressure W > 0 for an isolated system, (b) 0 W = Q = 0 E int = 0 Volume a f (c) Pressure 0 Pressure i iw > 0 Volume f f e can control how uch work it does. h Moving from f to i, it does negative work. Cycling clockwise yields Volume a positive net work. for a process (V i, P i ) to (V i, P i ), a cycle, E int = 0 i Q = W Pressure Pressure i W net > 0 Volume f d (e) 0 W < 0 Volume f (f) 0 Volume f

30 a system is always the negative of the work done by the Eq in terms of the work W on done on the system, on. This tells us the following:the internal energy of a e if heat Theis figure absorbed shows by four the system paths on or if a p-v positive diagram work along is which a gas nversely, canthe be taken internal from energy statetends i to to state decrease f. Rank if heat the paths, is greatest to egative least, work according is done ontothe system. (a) the change E int in the internal energy of the gas, Applying the 1st Law & P-V diagrams questions p ur paths on a p-v diagram taken from state i to state f. g to (a) the change E int in e gas, (b) the work W done gnitude of the energy transn the gas and its environi f A 1, 2, 3, 4 B 4, 3, 2, 1 int,are not true differentials;that is,there are no such functions as end only Con 1, the 4, state 3, 2of the system.the quantities dq and dw are are usually represented by the symbols d Q and d W. For our imply as Dinfinitesimally the same small energy transfers. 1 Based on question from Halliday, Resnick, and Walker, page 491. V

31 a system is always the negative of the work done by the Eq in terms of the work W on done on the system, on. This tells us the following:the internal energy of a e if heat Theis figure absorbed shows by four the system paths on or if a p-v positive diagram work along is which a gas nversely, canthe be taken internal from energy statetends i to to state decrease f. Rank if heat the paths, is greatest to egative least, work according is done ontothe system. (b) the work W done on the gas, Applying the 1st Law & P-V diagrams questions p ur paths on a p-v diagram taken from state i to state f. g to (a) the change E int in e gas, (b) the work W done gnitude of the energy transn the gas and its environi f A 1, 2, 3, 4 B 4, 3, 2, 1 int,are not true differentials;that is,there are no such functions as end only Con 1, the 4, state 3, 2of the system.the quantities dq and dw are are usually represented by the symbols d Q and d W. For our imply as Dinfinitesimally the same small energy transfers. 1 Based on question from Halliday, Resnick, and Walker, page 491. V

32 a system is always the negative of the work done by the Eq in terms of the work W on done on the system, on. This tells us the following:the internal energy of a e if heat Theis figure absorbed shows by four the system paths on or if a p-v positive diagram work along is which a gas nversely, canthe be taken internal from energy statetends i to to state decrease f. Rank if heat the paths, is greatest to egative least, work according is done ontothe system. (c) the energy transferred to the gas as heat Q. Applying the 1st Law & P-V diagrams questions p ur paths on a p-v diagram taken from state i to state f. g to (a) the change E int in e gas, (b) the work W done gnitude of the energy transn the gas and its environi f A 1, 2, 3, 4 B 4, 3, 2, 1 int,are not true differentials;that is,there are no such functions as end only Con 1, the 4, state 3, 2of the system.the quantities dq and dw are are usually represented by the symbols d Q and d W. For our imply as Dinfinitesimally the same small energy transfers. 1 Based on question from Halliday, Resnick, and Walker, page 491. V

33 Applying the 1st Law: new Vocabulary There are infinitely many paths (V i, P i ) to (V f, P f ) that we might consider. Ones that are of particular interest for modeling systems in engines, etc. are processes that keep one of the variabless constant. There is a technical name for each kind of process that keeps a variable constant.

34 Applying the 1st Law: new Vocabulary Adiabatic process is a process where no heat is transferred into or out of the system. Q = 0 This type of transformation occurs when the gas is in a thermally insulated container, or the process is very rapid, so there is no time for heat transfer. Since Q = 0: E int = W

35 Applying the 1st Law: new Vocabulary Isobaric process is a process that occurs at constant pressure. P i = P = P f This type of transformation occurs when the gas is free to expand or contract by coming into force equilibrium with a constant external environmental pressure. In this case, the expression for work simplifies: W = Vf V i P dv = P(V f V i )

36 Applying the 1st Law: new Vocabulary Isovolumetric process is a process that occurs at constant volume. V i = V = V f In an isovolumetric process the work done is zero, since the volume never changes. W = 0 E int = Q (An isovolumetric process can also be called an isochoric process.)

37 Applying the 1st Law: new Vocabulary Isothermal process is a process that occurs at constant temperature. T i = T = T f This kind of transformation is achieved by putting the system in thermal contact with a large constant-temperature reservoir. In an isothermal process, assuming no change of phase (staying an ideal gas!): E int = 0

38 Isothermal Expansion of an Ideal Gas Since T = 0, PV = nrt reduces to: PV = a where a is a constant. We could also write this as: 606 P = a VChapter 20 The First Law of Thermody Plotting this function: P i P f P i Isotherm PV = constant The curve is a hyperbola. f Isothermal Exp Suppose an ideal g This process is des hyperbola (see App stant indicates that Let s calculate th The work done on t process is quasi-stat V i V f V

39 Isothermal Expansion of an Ideal Gas Since nrt is constant, the work done on the gas in an isothermal expansion is: Vf W = V i Vf P dv nrt = V i V dv ( ) Vf = nrt ln V i ( ) Vi W = nrt ln V f

40 Questions Which path is isobaric? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

41 Questions Which path is isothermal? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

42 Questions Which path is isovolumetric? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

43 Questions Which path is adiabatic? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

44 Example 20.6: Boiling water This example illustrates how we can apply these ideas to liquids and solids also, and even around phase changes, as long as we are careful. Suppose 1.00 g of water vaporizes isobarically at atmospheric pressure ( Pa). Its volume in the liquid state is V i = V liq = 1.00 cm 3, and its volume in the vapor state is V f = V vap = 1671 cm 3. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the steam and the surrounding air; imagine that the steam simply pushes the surrounding air out of the way.

45 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V f = V vap = 1671 cm 3 L v = J/kg

46 Heat Transfer We are now changing gears. We are still thinking about heat in more detail, but we are not necessarily talking about ideal gases. (Section 20.7 of the textbook.)

47 Heat Transfer Mechanisms When objects are in thermal contact, heat is transferred from the hotter object to the cooler object There are various mechanisms by which this happens: conduction convection radiation

48 Conduction Heat can flow along a substance. When it does, heat is said to be transferred by conduction from one part of the substance to another. Some materials allow more heat to flow through them in a shorter time than others.

49 Conduction Heat can flow along a substance. When it does, heat is said to be transferred by conduction from one part of the substance to another. Some materials allow more heat to flow through them in a shorter time than others. These materials are called good conductors of heat: metals (copper, aluminum, etc)

50 Conduction In solids, conduction happens via vibrations collisions of molecules collective wavelike oscillations (phonons) diffusion and collisions of free electrons In liquids and gases, conduction happens through diffusion and collisions of molecules.

51 Conduction Some materials are not good conductors and are referred to as thermal insulators.

52 Conduction Some materials are not good conductors and are referred to as thermal insulators. Examples: air (and hence down feathers, wool) styrofoam wood snow

53 Newton s Law of Cooling (Applies for thermal conduction) Newton found a relation between the rate that an object cools and its temperature difference from its surroundings. Objects that are much hotter than their surroundings lose heat much faster than objects that are only a bit hotter than their surroundings. Using Q for heat: dq dt = ha T where A is the heat transfer surface area and h is the heat transfer coefficient

54 Newton s Law of Cooling (Applies for thermal conduction) dq dt = ha T If there is no phase change in the substance, then we can use the relation for heat capacity: Q = C T where in this case the heat is transferred out of the hot object to the environment. to get: d( T) = r T dt where the constant r = ha/c.

55 Newton s Law of Cooling Example Hot leftover soup must be cooled before it can be put in the refrigerator. To speed this process, you put the pot in a sink with cool, running water, that is maintained at 5 C. The hot soup cools from 75 C to 40 C in 8 minutes. Why might you predict that its temperature after another 8 minutes will be 22.5 C?

56 Summary more about phase changes work, heat, and the first law of thermodynamics P-V diagrams applying the first law in various cases heat transfer Homework Serway & Jewett: Read chapter 20 an look at the examples. prev: Ch 20, onward from page 615. OQs: 3, 5, 7; CQs: 3, 11; Probs: 1, 3, 9, 13, 19 new: Ch 20, OQs: 1, 13; Probs: 25, 27, 29, 31, 35, 39, 41, 59, 71, 77